ACSI 2021 Y3EXP FYE AMath P1 Solutions

2021 Y3 Express Additional Mathematics Final Examinations Paper 1 Solutions Solutions: 1 232 1xy y−+ = − -- (1) 21xy−= -- (2) From (2): 21yx= − -- (3) Sub (3) into (1): 23 (2 1) 2(2 1) 1xx x− −+ − = − 226 3 2(4 4 1) 1xx xx− ++ − += − 226 38 821x xx x− + + − += − 22 5 30xx− += ( 1)(2 3) 0xx− −= 10x−= or 2 30x−= 1x= or 3 2x= 1y= or 2y= Alternatively, From (2): 1 2 yx += -- (3) Sub (3) into (1): 2 22 13 212 3 34 2 y yy y yy +− += − − −+ = − 2 3 20yy− += ( 1)( 2) 0yy− −= 1y= or 2y= 1x= or 3 2x=

2 Solutions: 2(i) 8tan 15θ =− (ii) cos( ) cosθθ−= 15 17=− (iii) 1sec(180 ) cos(180 )θ θ°− = °− 1 cos( )θ= − 1 15 17 = −− 17 15= θ8 17 -15

3 Solutions: 3 Gradient 11 1 283 −= =− ( )1 23 5 or 1 2( 3) 25 cc YX YX = + ∴= − −= − ∴= − 2 2 2 11 25 25 yx x x = − −= 2 225 xy x∴= −

4 Solutions: 4(a) ( ) ( ) ( ) 2 22 22 3 22 2222 3 22 2 2 x x A Bx C xxxx x x A x Bx C x −+ + = + ++ − += + + + Let 0,x= 22 1 AA= ⇒= Compare coefficients of 2x : 3 2 12AB B= += + 1B∴= Compare coefficients of x : 22 C−= 1C∴= − ( ) 2 22 3 221 1 2222 xx x xxxx −+ − = + ++ (b) 2 22 45 2 143 43 xx xx xx ++ = +++ ++ 22 2 1 43 45 43 2 xxxx xx ++ ++ ++ 2 2 43 1 3 AB xx x x = +++ + + 2 ( 3) ( 1)Ax Bx= ++ + Let 1,x=− 22 1 AA= ⇒= Let 3,x=− 22 1 BB= −⇒= − 2 2 45 1 1 143 1 3 xx xx x x ++ = +−++ + +

5 Solutions: 5 (i) ( ) ( ) ( ) ( ) 32 15 11 f 2 11 22 2ab + + += = ( ) ( ) ( ) ( ) 32 15 14 f 1 14 11 1ab + + += −= −− − 2 15 11 ___(1) 84 82 8 44 ab ab ab ++ + = += − += − 1 15 14 ___(2)2 ab ab −+−+ = += (1) − (2): 36a =− 2 4( 2) 4 4 a b =− = −−−= (ii) 32 4 15 0 5 55 5f2 2 22 2 +++ = =−       (5 2 )x∴− is a factor of f ( ).x (iii) 2 32 32 2 2 23 2 5 2 4 15 25 44 4 10 - 6 15 - 6 15 0 xx x xx x xx xx xx x x ++ −+− ++ + + −+ −+ + + 2 2 f( ) 0 ( 25 ) ( 23 ) 0 25 0 o r 23 0 5 2 x x xx x xx x = −+ ++= − += + += = Method 1: Discriminant 22 4 2 4(1)(3) 80 b ac−= − = −< Method 2:Completing the Square ( ) 22 2 3 1 20xx x+ += + +> 2 23xx∴++ has no real roots and 5 2x= is the only real root.

6 Solutions: 6 (a) 2cosec 2 2x= 2 1 2sin 2 x = 2 1sin 2 2x= 1sin 2 2 x=± All quadrants 0 180x≤≤ ° 0 2 360x≤≤° 45α = ° 2 45 ,135 ,225 ,315x= °°°° 22.5 ,67.5 ,112.5 ,157.5x= ° °°° (b) (i) Amplitude 1 ( 5) 6=−− = 6a= Period is 8π 21 84b π π= = Axis: 5y=− 5c=−

7 Solutions: 7i 4322log 3 log 27 log ( 5)xx=+− 2 2 2 2log 3 3 log ( 5)log 4 x x= +− 22log 3 log ( 5) 3xx− −= 2 3log 3 5 x x =− 3 85 x x =− 3 8 40 5 40 xx x = − = 8x= ii ln(4 2) ln 2 ln 3 ln 2 ln 3 ln(2 3) x x x x+= + = + = ⋅ 4 223xx+= ⋅ ( ) 22 32 2 0xx− += Let 2x be y. 2 3 20yy− += ( 2)( 1) 0yy− −= 2 or 1yy= = 2 2 or 2 1xx= = 1 or 0xx= = iii lg 1 lg( 10)81 3 9 xx +−×= 4 lg 2 2lg( 10)33 3 xx +−×= 4 lg 2 2lg( 10)33 xx+ +−= 4 lg 2 2lg( 10)xx+= + − 2lg( 10) lg 2xx−−= 2lg( 10) lg 2xx−−= 2( 10)lg 2x x − = 2( 10) 100x x − = 2 20 100 100xx x−+= 2 120 100 0xx− += 119 or 0.839 (rejected 10 0)xx x= = −>

8 Solutions: 8 (i) Centre C ( ) 3 10 2,22 1,1 −+ += = − Radius 221 ( 1) (2 1) 5 = −− + − = (i) Gradient of perpendicular line 1 2=− [ ]11 ( 1)2 11 22 yx x −= − −− = −− 11 22yx= −+ Solve for intersection R. 1128 22 4 16 1 5 15 3 2 ( 3,2) xx xx x x yR += − + + = −+ =− = − ⇒= ∴− Alternatively, 22 22 22 22 ( 1) ( 1) 5 ( 1) (2 8 1) 5 2 1 4 28 49 5 5( 6 9) 0 ( 3) 0 3, 2 ( 3, 2) xy xx xx x x xx x xy R ++−= + + +− = + ++ + + = ++= ⇒+ = = − = ∴− (iii) Let centre of 2C be ( , ).xy 11, ( 3, 2)22 xy−+ = − 5, 3xy= −= Same radius of 5 Equation of 2C is 22( 5) ( 3) 5xy+ +− =